Preguntas de entrevistas para Production engineer

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Se le preguntó a Production Engineer…27 de mayo de 2015

Take a paragraph as Input and output the top three most repeated words

5 respuestas

from collections import Counter def get_most_3_repeated_words(paragraph): counter = Counter(paragraph.split()) return sorted(counter.items(), key=lambda x: x[1])[-3:] Menos

#/usr/bin/env python from collections import defaultdict wordlst = [] wordcnt = defaultdict(lambda : 0) paragraph = open('/home/nacho/python_test.txt','r') for line in paragraph.readlines(): for word in line.split( ): wordcnt[word] += 1 del wordlst paragraph.close() mostCommon3 = wordcnt.items() del wordcnt mostCommon3 = sorted(mostCommon3, key=lambda value: value[1], reverse=True) for indx in xrange(3): print "%s: %i" % (mostCommon3[indx][0], mostCommon3[indx][1]) Menos

A more correct shell answer, which returns the top word first: for word in `cat input.txt`; do echo $word; done | sort | uniq -c | sort -nr | head -3 | awk '{print $2}' Menos

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Meta

What does "$?" mean in bash?

5 respuestas

returns the status code of the last executed command

The exit code of the last executed command.

The exit code of the last executed command.

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Meta

What options do you have, nefarious or otherwise, to stop people on a wireless network you are also on (but have no admin rights to) from hogging bandwidth by streaming videos?

5 respuestas

This question is basically a way of ascertaining how much knowledge you have of the effect of distorting networking protocols in nonstandard ways. Everything from "ask them politely to stop" to "hijack their connection". Menos

set a dns server with the same MAC of real DNS server, forward all the "healthy" traffic back to real DNS, hijack the Domain related to video streaming. Menos

Anonymous on Apr 4, 2016: (but have no admin rights to)?

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Meta

For a given set of software checkins, write a program that will determine which part along the branch where the fault lies.

4 respuestas

A - array of versions, |A| = n check(A[0]) = true - first version is fine check(A[n-1]) = false - last version has the fault binary search: int start = 0; int end = n-1; while(end-start>1){ int middle = (end+start)/2; if (check(A[middle])) start = middle; else end = middle; } A[end] is the version with the fault Menos

first, design a test case to detect the failure. next, determine the last known "good" state of the software, i.e. the revision number of the last known "good" state. now, write a script that will check out each subsequent revision starting from the known good revision, build, run the test case and examine the output. When output fails, you can determine the revision at which the bug was introduced. Menos

I would write an integration test, for that scenario which started to fail. Then I'd write a shell script which would run that test over some amount of last revisions. Once test is failed, you've found a revision where mistake was made. Menos

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Meta

Why wouldn't you want a root DNS server to answer queries for you, instead of delegating you to an authoritative server?

4 respuestas

Because root DNS server can't handle zillions of requests at once, that's why you need a load balancer to split tasks on multiple DNS servers. every server has it's own limits, and load balancing helps to "scale" the limits. Menos

because it would be faster due also to caching mechanism.

DNS servers basically are text file doing Name to IP mapping(A type). So if a root server had to answer all the queries ,it would essentially be a million line text file, mapping each and every website.This would result in large CPU processing and time. Essentially, Root servers redirect the query to top level DNS servers and so on Menos

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Meta

Name three states a process can be in.

4 respuestas

Running, Stopped, IOwait

D = uninterruptible sleep (waiting for io) R= running S = interruptible sleep T= stopped by job control signal t= stopped by debugger with tracing x = dead z= zombie (dead) waiting for its return value to be recovered by parent process Source: ps man page Menos

NEW- The process is being created. READY- The process is waiting to be assigned to a processor. RUNNING- Instructions are being executed. WAITING- The process is waiting for some event to occur(such as an I/O completion or reception of a signal). TERMINATED- The process has finished execution. Menos

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AOL

You have two water buckets: the first can hold 3 gallons of water and the second can hold 5 gallons of water, how do you fill the 5 gallon bucket with exactly 4 gallons of water using these two buckets (the 3 gallon and 5 gallon bucket)?

4 respuestas

Easiest way: * Fill the 5 gallon bucket with water. * Pour out 3 gallons into the 3 gallon bucket, now there are 2 gallons of water in the 5 gallon bucket. * Empty the 3 gallon bucket. * Pour the remaining 2 gallons of water in the 5 gallon bucket into the 3 gallon bucket. * Refill the 5 gallon bucket with water again. * Fill the 3 gallon bucket with the 5 gallon bucket until full, 1 gallon will be removed from the 5 gallon bucket and thus you have 4 gallons of water in the 5 gallon bucket. Menos

There is another way of doing it. Take a full 3 gallon bucket and pour into the empty 5 gallon bucket. Take another full 3 gallon and pour it into the 5 gallon bucket until it gets full. Now there will be one gallon in the 3 gallon bucket. Now empty the 5 gallon bucket and pour the remaining one gallon from the 3 gallon bucket into the empty 5 gallon bucket. Now pour full 3 gallon bucket into the 5 gallon bucket.You will get 4 gallons in the 5 gallon bucket. Menos

Fill both buckets half way

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Flex

Why you chose this company

4 respuestas

I would like to gain more knowledge in the manufacturing field

Experiance job pvc production. Pvc fabrication pvc welder pvc bend making pvc coupler Menos

I would like to gain more knowledge in the manufacturing field

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OPPO

about work experience

3 respuestas

about machines

1 year experience and I worked as production engineer

Production engineer

Meta

Add each number in an array until the sum equals the rest of the array

3 respuestas

#!/usr/bin/env python3 # time complexity = O(n) arr = [1, -2, 3, 2, 0, -1, 10, -9, 2] arr_sum = 0 for i in range(0, len(arr)): arr_sum += arr[i] i_sum = 0 r_sum = arr_sum found = None for i in range(0, len(arr)): i_sum += arr[i] r_sum -= arr[i] if i_sum == r_sum: found = i break if found is None: print("no index found that satisfies the given requirement") else: print("array: {}".format(arr)) print("first: {}".format(arr[:found+1])) print("second: {}".format(arr[found+1:])) Menos

Out[17]: l = [1, -2, 3, 2, 0, -1, 10, -9, 2] In [16]: for i in xrange(len(l)): l_temp += l[i] if l_temp == sum(l[i+1:]): print "found index = {0}, value={1}".format(i, l[i]) break ....: out: found index = 5, value=-1 Menos

I dont know what is difficult about this question and why do they ask this from everyone Menos

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